DOMAIN AND RANGE OF COS INVERSE X

Domain of cosine inverse x :

For -1 ≤ x ≤ 1, define cos-1 x as the unique number y in [0, π] such that cos y = x.

In other words, the inverse cosine function

cos-1 : [-1, 1] --> [0, π]

is defined by cos-1(x) = y if only only if cos y = x and y ∈ [0, y]

Principal domain :

The restricted domain [0, π] is called the principal domain of cosine function and the values of y = cos-1(x), -1 ≤ x ≤ 1 are known as principal values of the function y = cos-1(x).

From the definition of  y = cos-1(x), we observe the following :

i) y =  cos-1(x), if and only if x = cos y for  -1 ≤ x ≤ 1 and  0 ≤ y ≤ π

ii) cos [cos-1(x)] = x, if |x| ≤ 1 and has no sense if |x| > 1

iii)  cos-1(cos x) = x, if ≤ x ≤ π, the range of cos-1 x.

Note that cos-1 (3π) = π

Find all values of x such that

Problem 1 :

-6π ≤ x ≤ 6π and cos x = 0

Solution :

cos x = 0

x = cos-1(0)

For what angle measure of cosine, we get the value 0.

General solution for cosine function for cos-1(0) :

Problem 2 :

-5π ≤ x ≤ 5π and cos x = 1

Solution :

cos x = 1

x = cos-1(1)

For what angle measure of cosine, we get the value 1.

General solution for cosine function for cos-1(1) :

x = 2nπ, where n ∈ N

 If n = 0x = 2(0)πx = 0 If n = 1x = 2(1)πx = 2π If n = -1x = 2(-1)πx = -2π If n = -2x = 2(-2)πx = -4π If n = 2x = 2(2)πx = 4π If n = -3x = 2(-3)πx = -6π

If n = 3

x = 2(3)π

x = 6π∉-5π ≤ x ≤ 5π

So, the possible values of n are

0, ±1, ±2, -3

Problem 3 :

State the reason for

Solution :

Here the angle measure (-π/6) mentioned is not in the domain [0, π]. So, it is not true.

Problem 4 :

Is cos-1(-x) = π - cos-1(x) true ? justify your answer ?

Solution :

cos-1(-x) = π - cos-1(x)

Here the value mention is negative. Using ASTC formula in 2n quadrant and in 3rd quadrant cosine will be negative. Considering the domain [0, π].

To get the angle, we should use

π - cos-1(x)

So, the given statement is true.

Problem 5 :

Find the principal value of cos-1 (1/2)

Solution :

cos-1 (1/2)

For what angle measure of cosine, we get 1/2.

For 60 degree, we get 1/2 and it lies in the domain [0, π].

cos-1 (1/2) = π/3

Find the value of

Problem 6 :

2cos-1(1/2) + sin-1(1/2)

Solution :

= 2cos-1(1/2) + sin-1(1/2)

= 2 (π/3) + π/6

= 2π/3 + π/6

= (4π + π)/6

= 5π/6

Problem 7 :

cos-1(1/2) + sin-1(-1)

Solution :

For what angle measure of cosine, we get 1/2 and for what angle measure of sin, we get -1.

=  (π/3) +  (3π/2)

Here  (π/3) is in the domain of cosine. But 3π/2 is not in the domain of sin.

So, -π/2 is the replacement of 3π/2.

=  (π/3) +  (-π/2)

= (2π - 3π)/6

= -π/6

Problem 8 :

Solution :

Problem 9 :

Solution :

Combining the above domain, we get [-5, 5]

So, the domain of the function f(x) is [-5, 5].

Problem 10 :

For what value of x, the inequality

π/2 < cos-1 (3x- 1) < π

holds ?

Solution :

To solve for x,

π/2 < cos-1 (3x- 1) < π

cos (π/2) < (3x - 1) < cos π

0 < 3x - 1 < -1

Add 1

1 < 3x < 0

Divide by 3

1/3 < x < 0/3

1/3 < x < 0

The inequality is true when,

0 < x < 1/3

Find the value of

Problem 10 :

Solution :

Problem 11 :

Solution :

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